How can I solve $y=a^x+x$ for x?

Question

I am looking for how to isolate x, or basically find the inverse function of $f(x)=a^x+x$. I don't have much experience with these types of equations, nor with non-elementary functions which may be necessary for this. I tried moving x over to the other side and taking a log, but that doesn't seem to bring me any closer to a solution.

Answer 1

$$y=a^x+x$$

Assuming $a>0$:

$$-x+y=e^{x \ln a}$$

Let $u=-x+y$ so that $x=-u+y$ and we get,

$$u=e^{(y-u) \ln a}=\frac{a^y}{e^{ u\ln a}}$$

So we have,

$$ue^{u \ln a}=a^y$$

$$u \ln a e^{u \ln a}= a^y\ln a$$

Now we utilize the lambert $W$ function which is defined by $W(x)e^{W(x)}=x$.

$$u \ln a=W(a^y \ln a)$$

$$u=\frac{W(a^y \ln a)}{\ln a}$$

$$x=y-\frac{W(a^y \ln a)}{\ln a}$$

Answer 2

Given $y$ you want $$(y-x)a^{y-x}=a^y$$

Letting $u=(y-x)\log a$ you get:

$$ue^u=a^y\log a$$

So $u=W(a^y\log a)$ and $$y-x=\frac{W(a^y\log a)}{\log a}$$

Or $$x=y-\frac{W(a^y\log a)}{\log a}$$

Here, $W$ is Lambert's W-function.